Table of Zonal Spherical Harmonics, 

 Hence we have 



Inside potential A= 8tt P + -j ^r 2 P 2 , 



Outside potential B= — • P + -^ ^^ ?2 ; 



519 



or 



B 1_ , 2 1„ a 



8^ = ? P » + 5? P2=fisay ' 



(1) 



(2) 



A« for our purpose, the actual unit of potential is unimportant, 



" ? A B 



we will use a for ^, and /3 for — 



I had no notion of how the equipotential surfaces would shape, 

 and I tried to avoid forming any such notion, as it was my 

 object to test the usefulness of the tables in working out any 

 new problem. The first thing that it strikes one to do in this 

 case is to find the potential on the sphere itself. This can be 

 done from either (1) or (2) by putting r=l, and then 



« = P + k ?2? 



or l+iP s . 



We can now find a for various values of 0, using the table. 

 Thus when = 0, P 2 = 1 , and therefore a = 1 4. Thus we have 

 the following values : — 











15 



30 



45 



60 



75 



90 



p 2 



a 



1 



1-4 



•9 



•625 



•25 



-•125 



-•3995 



-•5000 



1-36 



1-25 



1-1 



0-95 



0-84 



0-80 



Next, take any value of 0, say 0=45°. Then from the 

 table, P 2 = O2500, so that 



1 _ 



inside, a = 1 + — . 

 outside, 0=- +- ? , 



or 



10r' 2 + l 

 10 r 3 



For any value of r less than 1 we calculate a; for any value 

 of r greater than 1 we calculate /3. 



